3.2.44 \(\int \cot ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [144]

Optimal. Leaf size=81 \[ \frac {2 b \csc (c+d x)}{d}+\frac {a \csc ^2(c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d} \]

[Out]

2*b*csc(d*x+c)/d+a*csc(d*x+c)^2/d-1/3*b*csc(d*x+c)^3/d-1/4*a*csc(d*x+c)^4/d+a*ln(sin(d*x+c))/d+b*sin(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2800, 780} \begin {gather*} -\frac {a \csc ^4(c+d x)}{4 d}+\frac {a \csc ^2(c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d}+\frac {2 b \csc (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(2*b*Csc[c + d*x])/d + (a*Csc[c + d*x]^2)/d - (b*Csc[c + d*x]^3)/(3*d) - (a*Csc[c + d*x]^4)/(4*d) + (a*Log[Sin
[c + d*x]])/d + (b*Sin[c + d*x])/d

Rule 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+b \sin (c+d x)) \, dx &=\frac {\text {Subst}\left (\int \frac {(a+x) \left (b^2-x^2\right )^2}{x^5} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (1+\frac {a b^4}{x^5}+\frac {b^4}{x^4}-\frac {2 a b^2}{x^3}-\frac {2 b^2}{x^2}+\frac {a}{x}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {2 b \csc (c+d x)}{d}+\frac {a \csc ^2(c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 87, normalized size = 1.07 \begin {gather*} \frac {2 b \csc (c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d}+\frac {a \left (2 \cot ^2(c+d x)-\cot ^4(c+d x)+4 \log (\cos (c+d x))+4 \log (\tan (c+d x))\right )}{4 d}+\frac {b \sin (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(2*b*Csc[c + d*x])/d - (b*Csc[c + d*x]^3)/(3*d) + (a*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c + d*x]]
+ 4*Log[Tan[c + d*x]]))/(4*d) + (b*Sin[c + d*x])/d

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Maple [A]
time = 0.22, size = 101, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )}{d}\)	\(101\)
default	\(\frac {a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )}{d}\)	\(101\)
risch	\(-i a x -\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}+\frac {4 i \left (3 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-7 b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+7 b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/4*cot(d*x+c)^4+1/2*cot(d*x+c)^2+ln(sin(d*x+c)))+b*(-1/3/sin(d*x+c)^3*cos(d*x+c)^6+1/sin(d*x+c)*cos(
d*x+c)^6+(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)))

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Maxima [A]
time = 0.43, size = 69, normalized size = 0.85 \begin {gather*} \frac {12 \, a \log \left (\sin \left (d x + c\right )\right ) + 12 \, b \sin \left (d x + c\right ) + \frac {24 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 4 \, b \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*a*log(sin(d*x + c)) + 12*b*sin(d*x + c) + (24*b*sin(d*x + c)^3 + 12*a*sin(d*x + c)^2 - 4*b*sin(d*x +
c) - 3*a)/sin(d*x + c)^4)/d

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Fricas [A]
time = 0.38, size = 110, normalized size = 1.36 \begin {gather*} -\frac {12 \, a \cos \left (d x + c\right )^{2} - 12 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (3 \, b \cos \left (d x + c\right )^{4} - 12 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right ) - 9 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(12*a*cos(d*x + c)^2 - 12*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*sin(d*x + c)) - 4*(3*b*cos
(d*x + c)^4 - 12*b*cos(d*x + c)^2 + 8*b)*sin(d*x + c) - 9*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right ) \cot ^{5}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cot(c + d*x)**5, x)

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Giac [A]
time = 15.24, size = 82, normalized size = 1.01 \begin {gather*} \frac {12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, b \sin \left (d x + c\right ) - \frac {25 \, a \sin \left (d x + c\right )^{4} - 24 \, b \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 4 \, b \sin \left (d x + c\right ) + 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(12*a*log(abs(sin(d*x + c))) + 12*b*sin(d*x + c) - (25*a*sin(d*x + c)^4 - 24*b*sin(d*x + c)^3 - 12*a*sin(
d*x + c)^2 + 4*b*sin(d*x + c) + 3*a)/sin(d*x + c)^4)/d

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Mupad [B]
time = 6.67, size = 207, normalized size = 2.56 \begin {gather*} \frac {7\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {46\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {40\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(a + b*sin(c + d*x)),x)

[Out]

(7*b*tan(c/2 + (d*x)/2))/(8*d) + ((11*a*tan(c/2 + (d*x)/2)^2)/4 - (2*b*tan(c/2 + (d*x)/2))/3 - a/4 + 3*a*tan(c
/2 + (d*x)/2)^4 + (40*b*tan(c/2 + (d*x)/2)^3)/3 + 46*b*tan(c/2 + (d*x)/2)^5)/(d*(16*tan(c/2 + (d*x)/2)^4 + 16*
tan(c/2 + (d*x)/2)^6)) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d + (3*a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*tan(c/2
+ (d*x)/2)^4)/(64*d) - (b*tan(c/2 + (d*x)/2)^3)/(24*d) + (a*log(tan(c/2 + (d*x)/2)))/d

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